Given a three digit number $n$, let $f(n)$ be the sum of digits of $n$, their products in pairs, and the...
This is my first time posting so do correct me if I am doing anything wrong.
Please help me with this math problem from the British Maths Olympiad (1994 British Maths Olympiad1 Q1 Number Theory).
Starting with any three digit number $n$ (such as $n = 625$) we obtain a new number $f(n)$ which is equal to the sum of the three digits of $n$, their three products in pairs, and the product of all three digits.
Find all three digit numbers such that $frac{n}{f(n)}=1$.
The only solution I found is $199$, can someone verify it please?
elementary-number-theory contest-math
edited Nov 19 '18 at 12:31
amWhy
191k28224439
asked Nov 19 '18 at 5:36
3684
1277
|
show 2 more comments
This is my first time posting so do correct me if I am doing anything wrong.
Please help me with this math problem from the British Maths Olympiad (1994 British Maths Olympiad1 Q1 Number Theory).
Starting with any three digit number $n$ (such as $n = 625$) we obtain a new number $f(n)$ which is equal to the sum of the three digits of $n$, their three products in pairs, and the product of all three digits.
Find all three digit numbers such that $frac{n}{f(n)}=1$.
The only solution I found is $199$, can someone verify it please?
elementary-number-theory contest-math
edited Nov 19 '18 at 12:31
amWhy
191k28224439
asked Nov 19 '18 at 5:36
3684
1277
in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
– mathworker21
Nov 19 '18 at 5:44
So $f(199) = 19, 9918, 81$?
– steven gregory
Nov 19 '18 at 5:45
1
@mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
– 3684
Nov 19 '18 at 5:51
2
to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
– mathworker21
Nov 19 '18 at 5:56
1
@mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
– 3684
Nov 19 '18 at 6:00
|
show 2 more comments
This is my first time posting so do correct me if I am doing anything wrong.
Please help me with this math problem from the British Maths Olympiad (1994 British Maths Olympiad1 Q1 Number Theory).
Starting with any three digit number $n$ (such as $n = 625$) we obtain a new number $f(n)$ which is equal to the sum of the three digits of $n$, their three products in pairs, and the product of all three digits.
Find all three digit numbers such that $frac{n}{f(n)}=1$.
The only solution I found is $199$, can someone verify it please?
elementary-number-theory contest-math
edited Nov 19 '18 at 12:31
amWhy
191k28224439
asked Nov 19 '18 at 5:36
3684
1277
This is my first time posting so do correct me if I am doing anything wrong.
Please help me with this math problem from the British Maths Olympiad (1994 British Maths Olympiad1 Q1 Number Theory).
Starting with any three digit number $n$ (such as $n = 625$) we obtain a new number $f(n)$ which is equal to the sum of the three digits of $n$, their three products in pairs, and the product of all three digits.
Find all three digit numbers such that $frac{n}{f(n)}=1$.
The only solution I found is $199$, can someone verify it please?
elementary-number-theory contest-math
elementary-number-theory contest-math
edited Nov 19 '18 at 12:31
amWhy
191k28224439
asked Nov 19 '18 at 5:36
3684
1277
edited Nov 19 '18 at 12:31
amWhy
191k28224439
asked Nov 19 '18 at 5:36
3684
1277
edited Nov 19 '18 at 12:31
amWhy
191k28224439
edited Nov 19 '18 at 12:31
amWhy
191k28224439
edited Nov 19 '18 at 12:31
amWhy
191k28224439
191k28224439
asked Nov 19 '18 at 5:36
3684
1277
asked Nov 19 '18 at 5:36
3684
1277
asked Nov 19 '18 at 5:36
3684
1277
1277
in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
– mathworker21
Nov 19 '18 at 5:44
So $f(199) = 19, 9918, 81$?
– steven gregory
Nov 19 '18 at 5:45
1
@mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
– 3684
Nov 19 '18 at 5:51
2
to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
– mathworker21
Nov 19 '18 at 5:56
1
@mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
– 3684
Nov 19 '18 at 6:00
|
show 2 more comments
in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
– mathworker21
Nov 19 '18 at 5:44
So $f(199) = 19, 9918, 81$?
– steven gregory
Nov 19 '18 at 5:45
1
@mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
– 3684
Nov 19 '18 at 5:51
2
to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
– mathworker21
Nov 19 '18 at 5:56
1
@mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
– 3684
Nov 19 '18 at 6:00
in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
– mathworker21
Nov 19 '18 at 5:44
in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
– mathworker21
Nov 19 '18 at 5:44
So $f(199) = 19, 9918, 81$?
– steven gregory
Nov 19 '18 at 5:45
So $f(199) = 19, 9918, 81$?
– steven gregory
Nov 19 '18 at 5:45
1
1
@mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
– 3684
Nov 19 '18 at 5:51
@mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
– 3684
Nov 19 '18 at 5:51
2
2
to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
– mathworker21
Nov 19 '18 at 5:56
to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
– mathworker21
Nov 19 '18 at 5:56
1
1
@mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
– 3684
Nov 19 '18 at 6:00
@mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
– 3684
Nov 19 '18 at 6:00
|
show 2 more comments
2 Answers
2
active
oldest
votes
Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.
answered Nov 19 '18 at 6:07
user574848
16015
May I ask how you got to the solution so quick, do you just see the solution?
– 3684
Nov 19 '18 at 6:13
@3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
– user574848
Nov 19 '18 at 6:15
How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
– 3684
Nov 19 '18 at 6:19
add a comment |
Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:
We want to compute all possible integers $n$ such that $frac{n}{f(n)} = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write
$$n = 100a + 10b + c,$$
where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:
$$f(n) = abc + ab + bc + ac + a + b + c.$$
Now, in order to have $frac{n}{f(n)} = 1,$ we must have $n = f(n)$. This happens when
$$99a + 9b = abc + ab + bc + ac$$
$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$
Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by
$$boxed{{199, 299, 399, 499, 599, 699, 799, 899, 999}}$$
answered Nov 19 '18 at 6:08
Ekesh
5326
Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
– 3684
Nov 19 '18 at 6:15
It's okay. Hopefully my answer still helps you. I also used to participate in math competitions. I think that the best way to get faster is to just practice by doing many problems.
– Ekesh
Nov 19 '18 at 6:18
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004566%2fgiven-a-three-digit-number-n-let-fn-be-the-sum-of-digits-of-n-their-pr%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.
answered Nov 19 '18 at 6:07
user574848
16015
May I ask how you got to the solution so quick, do you just see the solution?
– 3684
Nov 19 '18 at 6:13
@3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
– user574848
Nov 19 '18 at 6:15
How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
– 3684
Nov 19 '18 at 6:19
add a comment |
Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.
answered Nov 19 '18 at 6:07
user574848
16015
May I ask how you got to the solution so quick, do you just see the solution?
– 3684
Nov 19 '18 at 6:13
@3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
– user574848
Nov 19 '18 at 6:15
How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
– 3684
Nov 19 '18 at 6:19
add a comment |
Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.
answered Nov 19 '18 at 6:07
user574848
16015
Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.
answered Nov 19 '18 at 6:07
user574848
16015
answered Nov 19 '18 at 6:07
user574848
16015
answered Nov 19 '18 at 6:07
user574848
16015
answered Nov 19 '18 at 6:07
user574848
16015
16015
May I ask how you got to the solution so quick, do you just see the solution?
– 3684
Nov 19 '18 at 6:13
@3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
– user574848
Nov 19 '18 at 6:15
How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
– 3684
Nov 19 '18 at 6:19
add a comment |
May I ask how you got to the solution so quick, do you just see the solution?
– 3684
Nov 19 '18 at 6:13
@3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
– user574848
Nov 19 '18 at 6:15
How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
– 3684
Nov 19 '18 at 6:19
May I ask how you got to the solution so quick, do you just see the solution?
– 3684
Nov 19 '18 at 6:13
May I ask how you got to the solution so quick, do you just see the solution?
– 3684
Nov 19 '18 at 6:13
@3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
– user574848
Nov 19 '18 at 6:15
@3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did
– user574848
Nov 19 '18 at 6:15
How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
– 3684
Nov 19 '18 at 6:19
How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight.
– 3684
Nov 19 '18 at 6:19
add a comment |
Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:
We want to compute all possible integers $n$ such that $frac{n}{f(n)} = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write
$$n = 100a + 10b + c,$$
where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:
$$f(n) = abc + ab + bc + ac + a + b + c.$$
Now, in order to have $frac{n}{f(n)} = 1,$ we must have $n = f(n)$. This happens when
$$99a + 9b = abc + ab + bc + ac$$
$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$
Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by
$$boxed{{199, 299, 399, 499, 599, 699, 799, 899, 999}}$$
answered Nov 19 '18 at 6:08
Ekesh
5326
Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
– 3684
Nov 19 '18 at 6:15
It's okay. Hopefully my answer still helps you. I also used to participate in math competitions. I think that the best way to get faster is to just practice by doing many problems.
– Ekesh
Nov 19 '18 at 6:18
add a comment |
Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:
We want to compute all possible integers $n$ such that $frac{n}{f(n)} = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write
$$n = 100a + 10b + c,$$
where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:
$$f(n) = abc + ab + bc + ac + a + b + c.$$
Now, in order to have $frac{n}{f(n)} = 1,$ we must have $n = f(n)$. This happens when
$$99a + 9b = abc + ab + bc + ac$$
$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$
Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by
$$boxed{{199, 299, 399, 499, 599, 699, 799, 899, 999}}$$
answered Nov 19 '18 at 6:08
Ekesh
5326
Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
– 3684
Nov 19 '18 at 6:15
It's okay. Hopefully my answer still helps you. I also used to participate in math competitions. I think that the best way to get faster is to just practice by doing many problems.
– Ekesh
Nov 19 '18 at 6:18
add a comment |
Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:
We want to compute all possible integers $n$ such that $frac{n}{f(n)} = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write
$$n = 100a + 10b + c,$$
where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:
$$f(n) = abc + ab + bc + ac + a + b + c.$$
Now, in order to have $frac{n}{f(n)} = 1,$ we must have $n = f(n)$. This happens when
$$99a + 9b = abc + ab + bc + ac$$
$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$
Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by
$$boxed{{199, 299, 399, 499, 599, 699, 799, 899, 999}}$$
answered Nov 19 '18 at 6:08
Ekesh
5326
Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:
We want to compute all possible integers $n$ such that $frac{n}{f(n)} = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write
$$n = 100a + 10b + c,$$
where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:
$$f(n) = abc + ab + bc + ac + a + b + c.$$
Now, in order to have $frac{n}{f(n)} = 1,$ we must have $n = f(n)$. This happens when
$$99a + 9b = abc + ab + bc + ac$$
$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$
Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by
$$boxed{{199, 299, 399, 499, 599, 699, 799, 899, 999}}$$
answered Nov 19 '18 at 6:08
Ekesh
5326
answered Nov 19 '18 at 6:08
Ekesh
5326
answered Nov 19 '18 at 6:08
Ekesh
5326
answered Nov 19 '18 at 6:08
Ekesh
5326
5326
Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
– 3684
Nov 19 '18 at 6:15
It's okay. Hopefully my answer still helps you. I also used to participate in math competitions. I think that the best way to get faster is to just practice by doing many problems.
– Ekesh
Nov 19 '18 at 6:18
add a comment |
Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
– 3684
Nov 19 '18 at 6:15
It's okay. Hopefully my answer still helps you. I also used to participate in math competitions. I think that the best way to get faster is to just practice by doing many problems.
– Ekesh
Nov 19 '18 at 6:18
Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
– 3684
Nov 19 '18 at 6:15
Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick?
– 3684
Nov 19 '18 at 6:15
It's okay. Hopefully my answer still helps you. I also used to participate in math competitions. I think that the best way to get faster is to just practice by doing many problems.
– Ekesh
Nov 19 '18 at 6:18
It's okay. Hopefully my answer still helps you. I also used to participate in math competitions. I think that the best way to get faster is to just practice by doing many problems.
– Ekesh
Nov 19 '18 at 6:18
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004566%2fgiven-a-three-digit-number-n-let-fn-be-the-sum-of-digits-of-n-their-pr%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$?
– mathworker21
Nov 19 '18 at 5:44
So $f(199) = 19, 9918, 81$?
– steven gregory
Nov 19 '18 at 5:45
1
@mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$
– 3684
Nov 19 '18 at 5:51
2
to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get
– mathworker21
Nov 19 '18 at 5:56
1
@mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time.
– 3684
Nov 19 '18 at 6:00